Question Number 69416 by ahmadshah last updated on 23/Sep/19
Commented by mr W last updated on 23/Sep/19
$${not}\:{defined}\:{for}\:{real}\:{numbers}! \\ $$$${domain}\:{of}\:{function}\:{x}^{\sqrt{\mathrm{2}}} \:{is}\:{x}\geqslant\mathrm{0}. \\ $$
Answered by mr W last updated on 23/Sep/19
$$−\mathrm{1}=−\mathrm{1}+\mathrm{0}{i}=\mathrm{cos}\:\pi+{i}\:\mathrm{sin}\:\pi={e}^{{i}\pi} \\ $$$$\left(−\mathrm{1}\right)^{\sqrt{\mathrm{2}}} ={e}^{{i}\sqrt{\mathrm{2}}\pi} =\mathrm{cos}\:\left(\sqrt{\mathrm{2}}\pi\right)+{i}\:\mathrm{sin}\:\left(\sqrt{\mathrm{2}}\pi\right) \\ $$$$=−\mathrm{0}.\mathrm{266255}−\mathrm{0}.\mathrm{963903}\:{i} \\ $$